Great question. I'm sure you learnt about the Big-O in that lesson, so I shall not elaborate on that. (My explanation will use O(N) which for the ease of explanation, can be assumed as the same as theta(N), but they are different!) The important part of Big-O (and theta) is...

python,c++,algorithm,recurrence

T(n) = 3T(n/3) + n/3 T(n/3) = 3T(n/9) + n/9 T(n) = 3(3T(n/9) + n/9) + n/3 = 9T(n/9) + 2*n/3 //statement 1 T(n/9)= 3T(n/27) + n/27 T(n) = 9 (3T(n/27)+n/27) + 2*n/3 // replacing T(n/9) in statement 1 = 27 T (n/27) + 3*(n/3) T(n) = 3^k* T(n/3^k) +...

date,momentjs,recurrence,moment-recur

Instead of this you can go with something as explained below with simple JS Date() object. 1. Find the range of dates and store them in an array. 2. Find the week of the first date. And the day using getDay(). 3. Now loop through all your dates and check...

By looking at your code it looks like you already have the structure in place just need to know how to utilize it. I'm going to slightly modify your if condition as shown below. for(var ind=0; ind<=between.length; ind++){ var thisDt = new Date(between[ind]); if (thisDt.getMonthWeek() === baseDtWk && thisDt.getDay() ===...

java,algorithm,dynamic-programming,recurrence

Not sure of your solution but here is an intuitive recursive approach with memoization. Let there be n books with ith book having pages[i] pages. Also let there be m subscribers. Also let dp[i][j] be the answer to problem if we were given only books i,i+1.....n and there are only...

algorithm,math,big-o,recurrence,master-theorem

One useful fact is that for any ε > 0, we know that log n = O(nε). We also know that log n = Ω(1) Let's see if this tells us anything. We know that your recurrence is bounded from above by this one for any ε > 0: S(n)...

algorithm,complexity-theory,recurrence

A generalization of the Master Theorem is the Akra-Bazzi method. Assuming that your O(n log n) is really Θ(n log n), we have g(x)=x log x, ai=1 for i=1 and i=2, b1=2/3 and b2=1/3. Then b1p+b2p=1 when p=1, g(u)/up+1=(log u)/u has integral (log²u)/2, and T(x) is Θ(x log²x)....

Okay, so in algorithm analysis, a recurrence relation is a function relating the amount of work needed to solve a problem of size n to that needed to solve smaller problems (this is closely related to its meaning in math). For example, consider a Fibonacci function below: Fib(a) { if(a==1...

dynamic-programming,pseudocode,recurrence

The +1 is because you need one extra coin. So for example, if you have: Vj = 5 Min[17] = 4 And you want to know the number of coins it will take to get 22, then the answer isn't 4, but 5. It takes 4 coins to get to...

javascript,date,datetime,recurrence

This function gives a date for every [interval] and [intervalType] (e.g. every 1 month) between two dates. It can also correct dates in weekends, if necessary. Is that what you had in mind? Here a jsFiddle demo. function recurringDates(startDate, endDate, interval, intervalType, noweekends) { intervalType = intervalType || 'Date'; var...

math,recurrence,recurrence-relation

First, we choose n = 2: T(n = 2) = (2^2 + 2 + 18) / 8 = 24 / 8 = 3 Great, that works. Now we know that there is a number k >= 2 that fulfills the definition of T(n). Then, let's set n = k+1 for...

django,date,notifications,recurrence,django-timezone

As others have said, Celery can schedule tasks to execute at a specific time. from celery.schedules import crontab from celery.task import periodic_task @periodic_task(run_every=crontab(hour=7, minute=30, day_of_week="mon")) def every_monday_morning(): print("This is run every Monday morning at 7:30") Install via pip install django-celery...

According to my calculation, the correct answer for n=3, k=8 should be 4, so your example should give you that. The program itself looks correct. Update: Here is how I do it on paper: We're starting from c(0) - since in this case the value is known in any case:...

algorithm,recursion,complexity-theory,recurrence

You can write T(n) = sqrt(n)⋅T(sqrt(n)) + sqrt(n) as T(n) = n1/2 + n3/4 + n7/8 + ... We know Σi=1,...,∞ 2-i = 1, so you can say T(n) = n1/2 + n3/4 + n7/8 + ... < n + n + n + ... Now you only have to...

algorithm,math,sum,series,recurrence

No one gave the mathematical approach, so I am adding the mathematical approach to this AP problem. Given series is 1k + 2k + 3k + .... + k.k(OR k^2) Therefore, it means that there are altogether k terms together in the given series. Next, as here all the consecutive...

algorithm,recurrence,recurrence-relation

For case (ii) we are finding a subset of {x1,..,xj} that uses xj and sums to i. However, we are making this sum by combining xj with a subset of {x1,..,x(j-1)}. If we had p(i-xj,j) then we would be saying that we can make a total of i by using...

data-structures,recurrence,master-theorem

This looks more like the Akra-Bazzi theorem: http://en.wikipedia.org/wiki/Akra%E2%80%93Bazzi_method#The_formula with k=1, h=0, g(n)=log n, a=(2)^{1/2}, b=1/2. In that case, p=1/2 and you need to evaluate the integral \int_1^x log(u)/u^{3/2} du. You can use integration by parts, or a symbolic integrator. Wolfram Alpha tells me the indefinite integral is -2(log u +...

I have solved the question. You won't get the recurrence if you doesn't get the event with something like this: $event = $service->events->get('primary', "recurringEventId"); Once you have $event, you can use: $preevent->recurrence[0] Thanks to Xeron for the answer....

c#,file,streamwriter,recurrence

You need to close the StreamWriter to force it to flush to disk. You also need to look in the current directory, which might not be what you think it is....

google-calendar,icalendar,recurrence

While it is part of RFC5545, RANGE based RECURRENCE-ID are poorly understood by most clients/servers, except maybe by Lotus Notes. What clients typically end up doing is to put an end to the old event (UNTIL parameter in the RRULE) and to create a new event (with a new UID)...

python,recursion,relationship,recurrence

Edit: Now using binary search on closest values along with memoization, Needs testing, but the idea is straightforward. This still relies on the fact that even and odd are increasing sequences. #!/usr/bin/env python # inspired by your code memo = {'even': {0: 1, 2: 2}, 'odd': {1: 1}} def f(k):...

algorithm,asymptotic-complexity,recurrence

If you try to apply the master theorem to T(n) = 2T(n/2) + n/log n You consider a = 2, b = 2 which means logb(a) = 1 Can you apply case 1?0 < c < logb(a) = 1. Is n/logn = O(n^c). No, because n/logn grow infinitely faster than...

dynamic-programming,recurrence,recurrence-relation

I don't really understand your recurrence relation: Let dp[i][j][k] represent sum up to i with j elements and k coins. I think you're on the right track, but I suggest simply dropping the middle dimension [j], and use dp[sum][coinsLeft] as follows: dp[0][0] = 1 // coins: 0, desired sum: 0...

math,recurrence,recurrence-relation

You're right that if sqrt is the ceiling of the square root, then you'll never reach 1 by repeatedly applying square roots. I'm going to assume that you meant to use the floor, which means that you will indeed eventually hit 1 as the recurrence unwinds. In this case, your...

algorithm,math,big-o,recurrence,recurrence-relation

Your claim that lg n * lg n = Θ(n) is false. Notice that the limit of (lg n)2 / n tends toward 0 as n goes to infinity. You can see this using l'Hopital's rule: limn → ∞ (lg n)2 / n = lim n → ∞ 2 lg...

algorithm,recursion,big-o,complexity-theory,recurrence

It looks like the lower bound is pretty good, so I tried to proof that the upper bound is O(log n / log log n). But let me first explain the other bounds (just for a better understanding). TL;DR T(n) is in Θ(log n / log log n). T(n) is...

algorithm,iteration,time-complexity,recurrence,upperbound

Look at step 4 T(n) = 2(2(2(2(2T(n-5) + 1) + 1) + 1) + 1) +1 ----- 4 T(n) = 2(2(2(2(2T(n-5))))) + 16 + 8 + 4 + 2 +1 = T(n) = 2^4* 2T(n-5) + 2^4 + 2^3 + 2^2 + 2^1 + 2^0 = T(n) = 2^4* 2T(n-5)...

algorithm,data-structures,mergesort,recurrence

I like to look at it as "runs", where the ith "run" is ALL the recursive steps with depth exactly i. In each such run, at most n elements are being processed (we will prove it soon), so the total complexity is bounded by O(n*MAX_DEPTH), now, MAX_DEPTH is logarithmic, as...

python,recursion,recurrence,recurrence-relation

Instead of trying to simplify this function mathematically, I simplified the algorithm in Python. As suggested by @LambdaFairy, I implemented memoization in the getNumberOfZombits(time) function. This optimization sped up the function a lot. Then, I passed to the next step, of trying to see what was the input to that...